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SCH4UI: Final Exam Review
UNIT: Organic Chemistry
1) Nomenclature
1. Make a summary table of organic functional groups using the heading: Name, Functional Group, IUPAC Suffix.
2. Name the following:
b.
c.
a.
d.
e.
f.
g.
i.
h.
3. Draw two different diagrams of the benzene molecule. Explain what each diagram shows, and state which diagram is
thought to be more correct.
2) Chemical Reactions
1. Explain why it is important to know Markovnikov’s rule when trying to predict the products of addition reactions to
alkenes and alkynes. Illustrate your explanation with an example.
2. The structure of the compound produced by this reaction:
is ____________________.
3. The primary reaction product of the reaction given below is ____________________.
4. Why does the hydrogenation of 2-butyne with 1.0 mole of hydrogen produce 2 isomers?
5. When benzene is reacted with liquid bromine, bromobenzene is formed.
a) Write the balanced chemical equation for this reaction.
b) What type of reaction is it?
6. When a straight chain alcohol is reacted with
,
, and
The structure of the original alcohol was _________________________.
are both formed.
7. When (CH3)3COH is reacted with an oxidizing agent, the major compound found in the reaction vessel afterward is
____________________.
8. The process of converting cyclohexanol to cyclohexanone is called ____________________.
9. The reverse process of esterification is called ____________________.
10. Write the structural formula of each compound.
a) an alcohol derived from butane
b) an aldehyde derived from butane
c) an acid derived from butane
11. Propanol may be oxidized to an aldehyde and then to an acid. Write the condensed structures and names of
the aldehyde and acid formed.
12. Write the condensed structure of the ester that is produced by the reaction of methanol and methanoic
acid. Name the ester.
13. The correct IUPAC name for the compound formed by the reaction below is _________________________.
14. Complete the following table.
Reaction
Type of reaction
15. What is the difference between addition polymerization and condensation polymerization? Give an example of a
polymer formed by each reaction.
UNIT: Atomic Theory and Periodic Trends
1) Periodic Table
1. Describe and explain the general trends in atomic radius in the periodic table.
2. Distinguish between the terms “ionization energy” and “electron affinity.”
3. What is the relationship between atomic radius and ionization energy? Explain your answer.
4. On the following outline of a periodic table, use arrows to indicate the directions of increasing
a) atomic radius
b) ionization energy
c) electron affinity.
2) Atomic Structure and Bonding
5. Complete the following table to describe the three intramolecular forces.
Force
ionic
covalent
metallic
Model/diagram
Nature of attraction
Energy (kJ/mol)
Example
6. Complete the following table to describe the six intermolecular forces.
Force
ion-dipole
Model/diagram
Nature of attraction
Energy (kJ/mol)
Example
hydrogen bond
dipole-dipole
dipole-induced
dipole
ion-induced dipole
dispersion
7. What types of intermolecular forces affect each compound or mixture? Explain your reasoning.
a) CH2Cl2
b) C5H12
c) O2 dissolved in H2O
d) NaCl dissolved in H2O
8. Create a table to summarize the differences in the physical properties of the different types of solids (atomic,
molecular, covalent network, ionic, and metallic).
3) Quantum Mechanics
1. What are the four quantum numbers in the quantum mechanical model of the atom? What do these numbers
represent?
2. Fill in the following outline of a periodic table to show the four energy sublevel (s, p, d, and f) blocks.
3. What are the allowed values of ml for an electron with each orbital-shape quantum number.
a) l = 3
b) l = 1
4. What are the allowed values of l for an electron with each principal quantum number.
a) n = 4
b) n = 6
5. Explain what is wrong with each set of quantum numbers.
a) n = 3, l = 3, ml = 2; name: 3d
b) n = 5, l = 3, ml = 5; name: 5f
6. Explain what is wrong with each set of quantum numbers.
a) n = 2, l = 1, ml = 0; name: 2p
b) n = 3, l = 2, ml = 2; name: 3d
7. Fill in the missing value(s) in each set of quantum numbers.
a) n = ?, l = 1, ml = 1; name: 3p
b) n = 1, l = ?, ml = ?; name: 1s
8. What is the name of the orbital that is represented by each set of quantum numbers?
a) n = 2, l = 1, m1 = 1
b) n = 4, l = 2, ml = 0
9. Explain what the aufbau principle is and how it is used.
10. Draw the electron configuration (energy level diagram) for oxygen and explain how you use Hund's Rule and the
Pauli Exclusion principle to do it.
11. Write the electron configuration for each element or ion.
a) Mn
b) Ca
c) Ag+
12. Write the condensed electron configuration for each element or ion.
a) Bi
b) Zn2+
c) Al
13. Draw the orbital diagram for each element or ion.
a) Co
b) Si
c) N
4) VSEPR Theory
1. For each molecule,
- draw the Lewis structure
- use the VSEPR theory to predict the shape of the compound
- decide whether or not the molecule is polar
- determine the bond angles
- draw the dipole movement for the molecule if the molecule is polar
a) NH4+
b) AsCl5
c) ICl
2. For each molecule,
- draw the Lewis structure
- use the VSEPR theory to predict the shape of the compound
- decide whether or not the molecule is polar
- determine the bond angles
- draw the dipole movement for the molecule if the molecule is polar
a) CH2Cl2
b) PH3
c) SeI6
UNIT: Energy and Rates of Reaction
1) Representing and Determining Energy Changes
1. Create a simple enthalpy diagram to represent the following reaction.
CaO(s)
(s)
2(s)
(g)
2. Create a simple enthalpy diagram to represent the following reaction.
Ag(s) + ½ Cl2(g) → AgCl(s)
3. The bond energy of H2(g) is 436.4 kJ/mol and for Cl2(g) it is 242.7 kJ/mol. The bond energy for the formation of HCl(g)
is 431.9 kJ/mol. What is the standard heat of formation for HCl (g) based on these values? What is the thermochemical
equation for this reaction?
4. 1.435 g of naphthalene (C10H8) is burned in a bomb calorimeter. There is exactly 2000.0 g of water surrounding the
naphthalene. The temperature of the water rises from 20.17 oC to 25.85oC. The heat capacity of the bomb calorimeter
is 1.80 kJ/oC.
a) What is the heat change for the burning of naphthalene?
b) What is the heat of combustion of naphthalene?
c) Write the thermochemical equation.
5. 0.412 g of calcium reacts in a coffee-cup calorimeter that contains 200.0 g of water. The initial temperature of the
water is 18.9oC, and the final temperature is 24.0 oC. Calculate the standard molar enthalpy change for this reaction.
6. 50.0 mL of 0.500 mol/L of copper(II)sulfate solution is mixed with 50.0 mL of 1.00 mol/L solution of sodium
hydroxide in a coffee-cup calorimeter. The temperature change is recorded to be 5.32 oC.
a) What is the enthalpy change for the reaction?
b) Write the thermochemical equation.
7. a) Given equations (1) and (2), calculate the enthalpy change for equation (3).
(1) Pb(s)  PbO2(s)  2SO3(g)  2PbSO4(s)
H  775 kJ
(2) SO3(g)  H2O(l)  H2SO4(aq)
H  133 kJ
(3) Pb(s)  PbO2(s)  2H2SO4(aq)  2PbSO4(s)  2H2O(l)
b) Draw an enthalpy diagram to represent equation (3).
8. Use standard heats of formation to calculate the heat of reaction for the following equation.
2H2S(g)
2(g) → 2H2O(l)
2(g)
2) Rates of Reactions and Reaction Mechanisms
1. The initial rate of the reaction: BrO3-(aq) + 5 Br-(aq) + 8 H+(aq)  3 Br2(l) + H2O(l)
Has been measured at the reactant concentrations shown (in mol/L)
Experiment
1
[BrO3-]
0.10
[Br-]
0.10
[H+]
0.10
Initial rate (mol/(L·s))
8.0  10-4
2
3
4
0.20
0.10
0.10
0.10
0.20
0.10
0.10
0.10
0.20
1.6  10-3
1.6  10-3
3.2  10-3
According to these results, what would be the initial rate (in mol/(L·s)) if all three concentrations are:
[BrO3-]=[Br-]=[H+]=0.20 mol/L?
2. Use the following diagram to answer the questions below.
a) Is the reaction exothermic or endothermic? Explain.
b) What letter represents the activation energy of the forward reaction?
c) What letter represents the heat of reaction?
d) What letter represents the activation energy of the reverse reaction?
3. The mechanism for a reaction is given below.
Step 1: A  B  C
Ea  168 kJ/mol
H  42 kJ/mol
Step 2: C  B  E  F
Ea  63 kJ/mol
H  21 kJ/mol
Step 3: F  B  G
Ea  84 kJ/mol
H  42 kJ/mol
a) Draw an accurate energy curve to represent the steps of
this reaction.
b) What is the overall equation for this reaction?
c) What is the H(forward) for the overall, or net, reaction?
d) Which step is the rate-determining step?
UNIT: Equilbrium
1) Introducing Equilibrium
1. Complete the following table, based on the following equilibrium system.
2Cl2(g)  2H2O(g)  4HCl(g)  O2(g) H  133 kJ
Stress
increase in temperature
increase in hydrogen chloride
increase in pressure at constant volume
decrease in volume
addition of inert gas at constant pressure
Equilibrium shift
2) Homogeneous Equilibrium Calculations
1. Name each equilibrium constant, and give an example. How do they relate?
a) Ka
b) Kb
c) Kc
d) Kw
e) Ksp
2. What is the equilibrium constant expression for the following equilibrium reaction?
H2O(g)  Cl2O(g)  2HOCl(g)
3. Determine the equilibrium constant for the following reaction, based on the equilibrium concentrations below.
3NO(g)  N2O(g)  NO2(g)
[NO]  0.060 mol/L
[N2O]  0.015 mol/L
[NO2]  0.025 mol/L
4. 0.150 mol of SO3 and 0.150 mol of NO are placed in a 1 L container and allowed to react as follows. At equilibrium,
the concentration of both SO2 and NO2 is 0.0621 mol/L. What is the equilibrium constant?
SO3(g)
(g)  SO2(g)
2(g)
5. The equilibrium constant for the following reaction is 14.3. If the equilibrium concentration of O 2 is 0.360 mol/L,
how much NO2 was introduced into the container?
2NO2(g) 2NO(g)
O2(g)
6. 0.440 mol/L of IBr is allowed to reach equilibrium, as shown below. The equilibrium constant is 6.62  103. What
are the equilibrium concentrations?
2IBr(g)  I2(g)  Br2(g)
7. N2(g) and O2(g) can exist in equilibrium with NO(g), as shown below. The equilibrium constant at 25.0oC is 4.8 x 10 -31.
If initially there are 1.25 mol and 0.50 mol of oxygen in a 1.00 L vessel, find the equilibrium concentrations of each
species.
N2(g)
2(g)  2NO(g)
3) Heterogeneous Equilibrium Calculations
1. Write the equilibrium reaction and the equilibrium expression for the dissolving of calcium phosphate.
2. Explain the common ion effect, using the example of sodium chloride being added to an equilibrium mixture
of silver chloride
3. The solubility of strontium sulfate at room temperature is 8.0  104 mol/L. Determine Ksp for strontium sulfate.
4. If the concentration of Ba2+ is found to be 4.2 x 10-8 mol/L in a saturated solution of this salt, what is the Ksp of
Ba3(PO4)2?
5. If 2.6 L of a saturated solution of MgCO3 is found to contain 1.1 g of MgCO3, what is the Ksp of MgCO3.
6. Calculate the concentrations of iron ions and hydroxide ions in a solution of iron(II) hydroxide (Ksp = 1.8  1015).
7. What is the molar solubility of barium fluoride, if Ksp for barium fluoride is 1.7  106?
8. 250 mL of 2.3  103 mol/L potassium iodate is reacted with an equal volume of 2.0  105 mol/L lead(II) nitrate. Will
a precipitate of lead(II) iodate (Ksp = 3.2  1013) form?
9. If 25 mL of 2.00  102 mol/L sodium hydroxide is added to 80 mL of 3.2  102 mol/L magnesium chloride, will a
precipitate form? Ksp for magnesium hydroxide is 1.2  1011.
10. What is the largest concentration of barium ions that can be added to a 2.3  102 mol/L solution of
sodium fluoride, without a precipitate of BaF2 forming? Ksp for barium fluoride is 1.7  106. (5)
3) Acid-Base Equilibrium
1. If the pH of an acid solution at 25oC is 6.18, what is the pOH; and the [H1+], [OH1-] in mol/L?
2. If the pOH of a base solution at 25oC is 5.67, what is the pH and the [H1+], [OH1-] in mol/L?
3. If the [H1+] of a solution at 25oC is 4.9 x 10-4 mol/L, calculate the [OH1-] in mol/L, the pH and the pOH.
4. Label the conjugate acid-base pairs in each reaction.
a) CO32(aq)  H2O(l)  HCO3(aq)  OH(aq)
b) H2SO4(aq)  H2O(l)  H3O+(aq)  HSO4(aq)
5. Explain the difference between a weak concentrated acid and a strong dilute acid.
6. List the following substances in order, from most acidic to least acidic: CH 4, H2O, HF, NH3. Explain your order.
7. Complete each acid equilibrium, and write the Ka expression.
a) H2PO3(aq)  H2O(l) 
b) CH3COOH(aq)  H2O(l) 
8. Complete each basic reaction, and write the Kb expression.
a) IO3(aq)  H2O(l) 
b) CH3NH2(aq)  H2O(l) 
9. What is the pH of a 1.47 mol/L solution of HCN(aq) if its Ka = 3.5 x 10-11?
10. What is the concentration of a monoprotic weak acid if its pH is 5.50 and its K a = 5.7 x 10-10?
11. What is the percent ionization of a 1.38 mol/L weak acid if its Ka = 2.7 x 10-6?
12. A solution of HF(aq) has an initial concentration of 0.075 mol/L. What are the equilibrium concentrations of HF (aq)
and its ions, if Ka  7.1  104?
13. What is the concentration of a weak base if its Kb = 1.4 x 10-11 and its pH = 8.75?
14. If the Kb of a 0.029 mol/L solution of weak base is 7.7 x 10 -7, what is its percent ionization?
15. Kb for caffeine, C8H10N4O2, is 4.1  104. If the equilibrium concentration of caffeine is 0.0250 mol/L, what was the
initial concentration of caffeine?
4) Titrations and Buffers
1. Distinguish between the end-point and the equivalence point in a titration.
2. 25 mL of standardized 0.45 mol/L NaOH is titrated with 21 mL of 0.35 mol/L acetic acid. Calculate the pH of the
solution.
3. 30.0 mL of a solution of a diprotic acid, oxalic acid (C 2H2O4), is titrated with 56 mL of a 0.050 mol/L
solution of potassium hydroxide. What was the concentration of the oxalic acid?
4. Sketch a graph that represents a titration of a weak acid with a strong base.
5. Sketch a graph that illustrates the change in pH as a weak base is titrated with a strong acid.
6. Predict the pH range of each solution if equal moles of each substance are titrated.
a) a strong acid neutralized with a weak base
b) a strong acid neutralized with a strong base
c) a weak acid neutralized with a strong base
SCH 4U Final Exam Review
Answer Section
1) Nomenclature:
1.
Name
Functional Group
IUPAC Suffix
alcohol
-ol
ether
-ether
aldehyde
-al
ketone
carboxylic acid
-one
-oic acid
ester
-oate
amine
-amine
amide
-amide
2. a. 2-methyl-1-butene
b. 3-methyl-2-butanol
c. 3-chloro-1-phenyl-1-propene
d. N-propylpropanamide e. 1-3-cyclopentadiene
f. 1-amino-3-methoxycyclohexane
g. 3-fluoro-3-nitrobutanoic acid h. 1-bromo-2-butene i. 4-ethoxypentanal
3.
The diagram with alternating single and double bonds implies that there are two different bond lengths and
strengths in benzene. The diagram with a circle in the middle shows that the electrons are delocalized and that all
the bonds are the same length. The diagram with the circle or dotted line, showing all six bonds to be equal, is
thought to be more correct.
2) Chemical Reactions
1. Markovnikov’s rule is important for an addition reaction between an asymmetrical alkene or alkyne and
another asymmetrical reactant, such as H2O or HBr. There are two possible products that may form as a result
of this addition. One product is the major product, and the other product is the minor product.
2.
4. - cis and trans isomers are produced:
- pi bond in double bond does not allow rotation
3.
5.
a) C6H6  Br2  C6H5Br  HBr
b) This is a substitution reaction.
6.
7.
(CH3)3COH
8.
9.
10.
oxidation
hydrolysis
a) CHOH
b) CHCHO
c) CHCOOH
11.
CH3CH2CHO, propanal; CH3CH2COOH, propanoic acid
12.
CHO–O–CH3, methyl methanoate
13.
cyclobutylmethanoate
14.
Reaction
Type of reaction
substitution
elimination
neutralization
hydrolysis
addition polymerization
15.
Addition polymerization is a reaction in which monomers with double bonds are joined together through a
series of addition reactions. For example, polyethylene is formed from ethene. Condensation polymerization
is a reaction in which monomers are joined together by the formation of ester or amide bonds and water is
released. Nylon-66 is an example of a condensation polymer.
UNIT: Atomic Theory and Periodic Trends
1) Periodic Table
1.
Atomic radius increases as you go down a group of elements in the periodic table. This trend is a result of
increasing numbers of electrons occupying increasing numbers of energy levels. The effective nuclear charge
changes only slightly and therefore does not offset the increase in size due to the increase in energy levels.
Atomic radius decreases as you go left to right across a period in the periodic table. The valence electrons
are found in orbitals of the same energy level. At the same time, the effective nuclear charge is increasing
with the increase in nuclear charge, which results in a greater force of attraction pulling the valence electrons
closer to the nucleus. Thus, atomic size decreases.
2.
Ionization energy is the energy that is required to completely remove one electron from a ground state
gaseous atom. Electron affinity is the change in energy that occurs when an electron is added to a gaseous
atom.
3.
Ionization energy is the energy that is required to remove an electron completely from a ground state gaseous
atom. This energy tends to increase as the atomic radius decreases. The closer the electrons are to the nucleus,
the great the force of attraction pulling or holding the electrons in the atom.
4.
2) Atomic Structure and Periodic Trends
1.
2.
3.
a) Dipole-dipole: The molecules of CH2Cl2 have a molecular dipole.
b) Dispersion forces: The C5H12 (hexane) molecule has covalent bonds between atoms.
c) Dipole-induced dipole: The H2O molecule has a bent shape, so it has a molecular dipole. This dipole
induces a dipole in the O2 molecules, resulting in their mutual attraction.
d) Ion-dipole interactions: NaCl dissociates into its ions as it dissolves in H2O. The H2O molecule has a
molecular dipole. Hence, ion-dipole interaction takes place between the ions of NaCl and the polar H2O
molecules.
4.
Type of Crystal
Solid
Atom
Molecular
Covalent
Network
Ionic
Boiling Point
low
generally low
(non-polar);
intermediate
polar
very high
high
Electrical Conductivity
in Liquid State
very low
very low
low
high
Other Physical Properties of
Crystals
very soft
non-polar: very soft; soluble in
non-polar solvents
polar: somewhat hard, but brittle;
many are soluble in water
hard crystals that are insoluble in
most liquids
hard and brittle; many dissolve in
Metallic
most high
water
all have a lustre, are malleable
and ductile, and are good
conductors; they dissolve in other
metals to form alloys
very high
3) Quantum Mechanics
1.
1) The principal quantum number, n, indicates the energy level of an atomic orbital and its relative size.
2) The orbital-shape quantum number, l, indicates the shape of the orbital.
3) The magnetic quantum number, ml, indicates the orientation of the orbital.
4) The spin quantum number, ms, indicates the direction in which the electron is spinning.
2.
3.
a) ml = 3, 2, 1, 0 1, 2, 3
b) ml = 1, 0, 1
4.
a) l = 0, 1, 2, 3
b) l = 0, 1, 2, 3, 4, 5
5.
a) l cannot equal 3, since the maximum value of l is (n  1) and d orbitals have an l value of 2.
b) ml cannot equal 5, since the maximum value of ml is +l.
6.
a) l cannot equal 1, since values of l can only be positive integers.
b) This set of quantum numbers is a valid set.
7.
a) n = 3
b) l = 0, ml = 0
8.
a) 2p
b) 4d
9.
The aufbau principle is the imaginary process of building up the ground state electron structure for each atom,
in order of atomic number. When determining the electron configuration of an element, the electrons are
written sequentially in orbitals of increasing energy, starting with the electron in the 1s orbital.
10.
1s22s22p4
Pauli exclusion principle states that no two electrons can have the same four quantum numbers, therefore each
orbital can hold only two electrons with opposite spins. Hund's rule says that the electrons in orbitals with the
same energy are half filled first before more are added. Also, the electrons in those half filled orbitals must
have the same spin.
11.
a) 1s2 2s2 2p6 3s2 3p6 4s2 3d5
c) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 5s2 4d8
b) 1s2 2s2 2p6 3s2 3p6 4s2
12.
a) [Xe] 6s2 4f14 5d1 6p3
c) [Ne] 3s2 3p1
b) [Ar] 4s2 3d10
13.
4) VSEPR Theory
1.
2.
UNIT: Energy and Rates of Reaction
1) Representing and Determining Energy Changes
1.
3. Hf  Bonds broken Bonds formed
 [ (Bond energy of H2(g))  (Bond energy Cl2(g))]  [Bond energy HCl(g)]
 [ (436.4) 
(242.7)]  [431.9]
 92.4 kJ/mol HCl(g)
The enthalpy of formation of HCl(g) is 92.4 kJ/mol HCl.
H2(g)  Cl2(g)  HCl(g)  92.4 kJ
4.
b)
Qlost by naphthalene  –[Qgained by water  Qgained by calorimeter]
 –[mcT  CT]
 –[(2000.0 g)(4.184 J/gC)(25.85C  20.17C)
 (1.80 kJ/C)(25.85C – 20.17C)]
Qlost by naphthalene  –[47 530 J  10.2 kJ]
 –47.5 kJ  10.2 kJ]
 –57.7 kJ
The heat change for the burning of naphthalene is –57.7 kJ.
Qlost by naphthalene  nHf
1.435 g  128.18 g/mol  0.011 20 mol
57.7 kJ  0.011 20 mol  Hf
Hf  –5.15  103 kJ/mol
The heat of combustion of naphthalene is –5.15  103 kJ/mol.
c) C10H8(l)  12O2(g)  10CO2(g)  4H2O(l)  5.15  103 kJ/mol
5.
Qlost by calcium  Qgained by water
nHr  mcT
 0.0103 mol
0.0103 mol  Hr  200.0 g  4.184 J/gC  5.1C
Hr  4.14  105 J/mol
The enthalpy change for the reaction of calcium in water is 4.14  102 kJ/mol Ca.
6.
a)
Qreaction  Qgained by solution
nHr  mcT
0.025 mol  Hr  100.0 g  4.184 J/gC  5.32C
Hr  8.9  105 J/mol
The enthalpy change for this reaction is 89.0 kJ/mol CuSO4.
b) CuSO4(aq)  2NaOH(aq)  Cu(OH)2(s)  Na2SO4(aq)  89 kJ
7.
a) Required equation: Pb(s) + PbO2(s) + 2H2SO4(aq)  2PbSO4(s) + 2H2O(l)
Pb(s)  PbO2(s)  2SO3(g)  2PbSO4(s)
H  775 kJ
2H2SO4(aq)  2SO3(g)  2H2O(l)
H  +266 kJ
Pb(s)  PbO2(s)  2H2SO4(aq)  2PbSO4(s) + 2H2O(l)
H  509 kJ
The enthalpy change for equation (3) is 509 kJ/mol.
b)
8. Hrxn  [2(Hf SO2(g)  2(Hf H2O(l))]  [2(Hf H2S(g))  3(Hf O2(g))]
 [2(296.8 kJ/mol)  2(285.8 kJ/mol)]  [2(20.6 kJ/mol)  3(0.0 kJ/mol)]
 1124 kJ/2 mol H2S(g)
562 kJ/mol H2S(g)
The heat of reaction is 562 kJ/mol H2S(g).
Rates of Reactions and Reaction Mechanisms
1. Using Experiment 1 and 2, it is evident that the rate doubles when [BrO3-] doubles, then this reaction is first order
with respect to [BrO3-]
Using Experiment 1 and 3, it is evident that the rate doubles when [Br -] doubles, then this reaction is first order with
respect to [Br-]
Using experiment 1 and 4, it is evident that the rate quadruples when the [H +] doubles, then this reaction is second order
with respect to [H+]
From experiment 1
rate = k [BrO3-][Br-][H+]2
8.0  10-4= k(0.1)(0.1)(0.1)2
k= 8
When all concentrations are 0.20 mol/L
rate = 8(0.20)(0.20)(0.20)2
=1.28  10-2 mol/(L·s)
The rate of the reaction would be 1.28 x 10 -2 mol/(L·s).
2. The reaction is endothermic. The enthalpy of the products is higher than the enthalpy of the reactants. This means that
energy is absorbed as the reaction proceeds.
b) A
c) C
d) B
3. a)
b) A  3B  E  G
c) Hr  H1  H2  H3
 Hr  42  21  42  21 kJ/mol
d) The rate-determining step is the first step because it has the highest activation energy barrier.
UNIT: Equilibrium
1) Introducing Equilibrium
1.
Stress
increase in temperature
increase in hydrogen chloride
increase in pressure at constant volume
decrease in volume
addition of inert gas at constant pressure
Equilibrium shift
to the right
to the left
to the left
to the left
to the right
2) Homogeneous Equilibrium Calculations
1.
Ka is the equilibrium constant for an acid. Example: Ka =
Kb is the equilibrium constant for a base. Example: Kb =
Kc is the equilibrium constant based on concentrations. Example: Kc =
Kw is the equilibrium constant for water. Example: Kw = [H3O+(aq)][OH(aq)]
Ksp is the equilibrium constant for an ionic solid. Example: Ksp = [Ag+(aq)][Cl(aq)]
2.
Keq 
3.
Keq 

 1.74
4.
Concentration (mol/L)
Initial
Change
Final
SO3(g)
0.150
0.0621
0.0879

NO(g)
0.150
0.0621
0.0879

SO2(g)
0.0
0.0621
0.0621

NO2(g)
0.0
0.0621
0.0621
Keq 

 4.99  101
The equilibrium constant is 4.99  101.
5.
Concentration (mol/L)
Initial
Change
Final
2NO2(g)
y
0.720
y  0.720

2NO(g)
0.0
2(0.360) = 0.720
0.720
+
O2(g)
0.0
0.360
0.360
Keq 

 14.3
(y  0.720)2 
(y  0.720)2  1.31  10–2
Take the square root of both sides.
y  0.720  0.114
y 0.834
Therefore, 0.834 mol of NO2 was introduced for each litre of volume.
6.
Concentration (mol/L) 2IBr(g) 
I2(g)

Br2(g)
Initial
0.440
0.0
0.0
Change
2x
x
x
Final
0.440  2x
x
x
Keq 

 6.62  103
Since the expression is a perfect square, take the square root of both sides.
 0.0814
x
x
1.163x 
x
0.0814(0.440  2x)
0.0358  0.163x
0.0358
0.030
The equilibrium concentrations of I2(g) and Br2(g) are both 0.030 mol/L. The equilibrium concentration of IBr
is 0.440  2(0.030)  0.38 mol/L.
7.
Concentration (mol/L)
Initial
Change
Final
N2(g)
1.25
x
1.25  x

O2(g)
0.50
x
0.50  x

2NO(g)
0.0
2x
2x
Keq 

 4.8  1031
Since the equilibrium constant is so small, the concentrations of the products will be extremely small.
Therefore, 1.25  x is approximately equal to 1.25, and 0.50  x is approximately equal to 0.50. Since the
expression is a perfect square, take the square root of both sides.
31
 4.8  10
(2x)2  3.0  1031
2x  5.48  1016
x  2.74  1016
The equilibrium concentration of NO(g) is 2(2.74  1016)  5.48  1016 mol/L. The equilibrium concentration
of N2(g) is 1.25 mol/L, and the equilibrium concentration of O2(g) is 0.50 mol/L.
3) Heterogeneous Equilibrium Calculations
1.
ANS:
Ca3(PO4)2(s)  3Ca2+(aq) + 2PO43 (aq)
Ksp = [Ca2+(aq)]3[PO43(aq)]2
2.
The common ion effect is observed when an ion is present in two different compounds that are mixed in the
same solution to produce an equilibrium condition. Adding sodium chloride to an equilibrium mixture of
silver chloride causes more silver chloride to precipitate out of solution, since the ion product constant is
affected.
3.
SrSO4(s)  Sr2+(aq) + SO42(aq)
Ksp = [Sr2+(aq)][SO42(aq)]
[Sr2+(aq)] = [SO42(aq)] = [SrSO4(s)] = 8.0  104 mol/L
Ksp = [Sr2+(aq)][SO42(aq)]
= (8.0  104 mol/L)(8.0  104 mol/L)
= 6.4  107
Therefore, Ksp for strontium sulfate is 6.4  107.
4.
Ba3(PO4)2(s) <====>
@E
3 Ba2+(aq)
4.2  10-8
+
2 PO43-(aq)
2/3  4.2  10-8
Ksp = (4.2  10-8 )3 ( 2/3  4.2  10-8 )2 = 5.8  10-38
5.
MgCO3(s) <====>
@E
Mg2+(aq)
1.1 g / 84.314 g/mol / 2.6 L
+
CO32-(aq)
1.1 g / 84.314 g/mol / 2.6 L
Ksp = (1.1 g / 84.314 g/mol / 2.6 L )2 = 2.5  10-5
6.
Fe(OH)2(s)  Fe2+(aq) + 2OH(aq)
Ksp = [Fe2+(aq)][OH(aq)]2
Concentration (mol/L)
Initial
Change
Equilibrium
Ksp
Ksp
1.8  1015
x
[Fe2+(aq)]
[OH(aq)]
Fe(OH)2(s)  Fe2+(aq) +
---0
---x
---x
2OH(aq)
0
2x
2x
= [Fe2+(aq)][OH(aq)]2
= (x)(2x)2
= 4x3
= 7.66  106 mol/L
= 7.66  106 mol/L
= 2(7.66  106 mol/L)
= 1.53  105 mol/L
7.
BaF2(s)  Ba2+(aq) + 2F(aq)
Ksp = [Ba2+(aq)][F(aq)]2
Concentration (mol/L)
Initial
Change
Equilibrium
BaF2(s) 
----------
Ba2+(aq) +
0
x
x
2F(aq)
0
2x
2x
Ksp = [Ba+(aq)][F(aq)]2
Ksp = (x)(2x)2
1.7  106 = 4x3
x = 7.5  103 mol/L
The molar solubility of barium fluoride is 7.5  103 mol/L.
8.
Pb(IO3)2(s)  Pb2+(aq) + 2IO3(aq)
[Pb2+(aq)] = [Pb(NO3)2(aq)] 
= (2.0  105 mol/L) 
= 1.0  105 mol/L
[IO3(aq)] = [KIO3(aq)] 
= (2.3  103 mol/L) 
= 1.15  103 mol/L
Qsp = [Pb (aq)][IO3(aq)]2
= (1.0  105 mol/L)(1.15  103 mol/L)2
= 1.32  1011
Since Qsp > Ksp, Pb(IO3)2 will precipitate until Qsp = 3.2  1013.
2+
9.
ANS:
Mg(OH)2(s)  Mg2+(aq) + 2OH(aq)
[Mg2+(aq)] = [MgCl2(aq)] 
= (3.2  102mol/L) 
= 2.4  102 mol/L
[OH(aq)] = [NaOH(aq)] 
= (2.00  102 mol/L) 
= 4.8  103 mol/L
Qsp = [Mg2+(aq)][OH(aq)]2
= (2.4  102 mol/L)(4.8  103 mol/L)2
= 5.5  107
Since Qsp > Ksp, Mg(OH)2 will precipitate until Qsp = 5.5  107.
10.
BaF2(s)  Ba2+(aq) + 2F(aq)
This problem involves the common ion effect.
Ksp = [Ba2+(aq)][F-(aq)]2
Concentration (mol/L)
Initial
Change
Equilibrium
BaF2(s) 
----------
Ba2+(aq) +
0
x
x
2F(aq)
2.3  102
x
(2.3  102) + x
Since x is small, use an approximation.
(2.3  102) + x  2.3  102
Ksp = [Ba2+(aq)][F(aq)]2
Ksp = (x)(2.3  102)2
9.0  109 = (x)(2.3  102)2
x = 1.70  105 mol/L
The largest concentration of barium ions that can be added to a 2.3  102 mol/L solution of fluoride ions,
without a precipitate of BaF2 forming, is 1.70  105 mol/L.
3) Acid/Base Equilibrium
1.
pOH = 7.82, [H1+] = 6.6  10-7 mol/L., [OH1-] = 1. 5  10-8 mol/L.
2.
pH = 8.33, [H1+] = 4.7  10-9 mol/L., [OH1-] = 2.1  10-6 mol/L.
3.
[OH1-] = 2.0  10-11 mol/L, pH = 3.31, pOH = 10.69
4.
a)
b)
CO32(aq)
base 1
H2SO4(aq)
acid 1


H2O(l)
acid 2
H2O(l)
base 2


HCO3(aq)
conjugate acid 1
H3O+(aq)
conjugate acid 2


OH(aq)
conjugate base 2
HSO4(aq)
conjugate base 1
5.
- A concentrated acid has lots of solute particles in a given volume. A dilute acid has a small number of
solute particles in a given volume. A strong acid is completely ionized. A weak acid is only partially
ionized.
- A weak concentrated acid has a high concentration of undissociated particles. A strong dilute acid has only
a small number of dissociated particles and no undissociated particles.
6.
HF, H2O, NH3, CH4
The strength of binary acids is related to the electronegativity of the element that is bonded to the hydrogen.
Here the order of decreasing electronegativity is F, O, N, C.
7.
a) H2PO3(aq)  H2O(l)  HPO32(aq)  H3O+(aq)
Ka 
b) CH3COOH(aq)  H2O(l)  CH3COO(aq)  H3O+(aq)
Ka 
8.
a) IO3(aq)  H2O(l)  HIO3(aq)  OH(aq)
Kb 
b) CH3NH2(aq)  H2O(l)  CH3NH3+(aq)  OH(aq)
Kb 
9.
initial
shift
@E
HCN(aq) <====> H1+(aq) + CN1-(aq)
1.47
-x
x
x
x
x
1.47  x
x2 / 1.47 = 3.5  10-11
x = (3.5  10-11  1.47)0.5
pH = – log x
pH = 4.56
pH = 5.14
10.
HX(aq) <=====> H1+(aq) + X1-(aq)
x
10-pH
10-pH
[H1+] = 10-pH = 10-5.50
[H1+] = 3.16  10-6 mol/L
(3.16  10-6)2 /  = 5.7  10-10
x = 1.8  10-2 mol/L
11.
initial
shift
@E
HY(aq) <====> H1+(aq) + Y1-(aq)
1.38
-x
1.38  x
x
x
x
x
x2 / 1.38 = 2.7  10-6
x = (2.7  10-6  1.38)0.5
pH = – log x
pH = 2.71,
% Ionization = 100(x / 1.38) = 0.14%
12.
Concentration (mol/L)
Initial
Change
Final
HF(aq)
0.075
x
0.075  x

H2O(l)

H3O+(aq)
0.0
x
x

F(aq)
0.0
x
x
Ka 

 7.1  104
x2  (0.075  x)(7.1  104)
x2  (5.325  105)  (7.1  104)x
x2 + (7.1  104)x  (5.325  105)  0
Use the general formula for a quadratic equation.
x  6.9  103
The equilibrium concentrations of H3O+(aq) and F(aq) are 6.9  103 mol/L. The equilibrium concentration of
HF(aq) is 0.075  (6.9  103) = 0.068 mol/L.
13.
XOH(aq) <=====> X1+(aq) + OH1-(aq)
x
10-pOH
10-pOH
pOH = 14 – pH = 14 – 8.75 = 5.25
[OH1-] = 10-pOH = 10-5.25
[OH1-] = 5.6  10-6 mol/L
(5.6  10-6)2 /  = 1.4  10-11 mol/L
x = (5.6  10-6)2 / 1.4  10-11 = 2.3 mol/L
14.
initial
shift
@E
XOH <=====> X1+ + OH10.029
-x
x
x
0.029
x
x
x2 / 0.029 = 7.7  10-7
x = 1.5  10-4 = [OH1-]
% ionized = 100  (1.5  10-4 / 0.029)
0.52 % ionized
15.
Concentration (mol/L)
Initial
Change
Final
C8H10N4O2(aq)
0.0250 + x
x
0.0250
+
H2O(l)

C8H11N4O2+(aq)
0.0
x
x
+ OH(aq)
0.0
x
x
Kb 

 4.1  104
x2  1.025  105
x  3.2  103
The initial concentration of caffeine was (2.50  102)  (3.2  103)  2.82  102 mol/L.
4) Titrations and Buffers
1.
The end-point is the point at which the indicator changes colour. The equivalence point is the point at which
just enough acid and base have been mixed for a complete reaction to occur, with no excess of either reactant.
2.
ANS:
N=CV
(mol)
initial
reacted
left over
NaOH
(0.45 mol/L  0.025 L)
0.01125
0.00735
0.0039
+
HC2H3O2 -----> NaC2H3O2 + H2O
(0.35 mol/L  0.021 L)
0.00735
0.00735
0
COH1- = n/V = 0.0039 mol / 0.046 L = 0.0848 mol/L
pOH = –log[0.0848] = 1.07
pH = 14 – 1.07 = 12.93
3.
2KOH(aq) + C2H2O4(aq)  K2C2O4(aq) + 2H2O(l)
Moles KOH = C  V
= 0.056 L  0.050 mol/L
= 2.8  103 mol
Moles C2H2O4 = 2.8  103 mol KOH 
= 1.4  103 mol
= 1.4 
= 4.7  102 mol/L
The concentration of the oxalic acid was 4.7  102 mol/L.
4.
5.
6.
a) pH < 7
b) pH = 7
c) pH > 7